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The concept of entropy is very ubiquitous, we learn about its uses starting from Information Theory (Shannon entropy) up to its basic definition in statistical mechanics in terms of number of micro-states.

Limiting the discussion to physics, when studying a physical system, can be a box filled with an ideal gas, a melt of polymers or the state of rods/molecules in a liquid crystalline system, in all such scenarios, there are specific entropic terms that we define in describing the evolution of the system (by including it in the free energy expression).

From a statistical mechanics point of view, we use Boltzmann's definition of: $$S=k_B \ln\Omega$$ where $\Omega$, can be the partition function in an ensemble, or simply the number of microstates within a given macrostate. But of course we almost never use this exact form of entropy in studying real systems, as it is impossible to count the microstates by any means. Instead we define entropic terms based on macroscopic variables of a system, like the followings (examples among the usual ones):

  • For a perfect gas one can write the entropy per atom of N atoms in volume V as: $$S_{\rm ideal}=k_B \ln\left(a\frac{V}{N}\right)$$ with $a$ a constant.

  • In soft condensed matter, studying liquid crystallinity, often we define an orientational entropy, describing the entropy lost when molecules are oriented. In its most general form defined as: $$S_{\rm orient}=-k_B \int f(\theta)\ln f(\theta)d\Omega$$ where $f(\theta)$ is an orientation distribution function and $d\Omega$ a small solid angle.

  • In polymers physics, often there are entropic terms attributed to homogeneity in number density distribution of monomers $n(i)$ along the chains (often called Lifshitz entropy), described as $$S_{\rm homogeneity}\propto -\left(\nabla \sqrt{n(i)}\right)^2$$ which is just the gradient of a density distribution.

In all such cases, it is relatively straightforward to see as to why we refer to such state functions as "entropic", as they all boil down to describing a certain kind of disorder in the system, and how their inclusion would effect the equilibrium state of a system. But the underlying question is about, how we give physical and mathematical bounds to these entropic terms must be consistent, hence there must be a set of axioms that a function must fulfill in order to qualify as an entropic term.

To elaborate on the question:

On the one hand, from a physical point of view, the entropy is always set to reach its maximum at the most disordered state, and a minimum for the most ordered case (i.e. 1 possible micro-state, complete certainty over its state). For example in liquid crystals again, we want $S_{\rm orient}$ to be at maximum in the orientationally disordered state, the isotropic, and for it to be at minimum in the completely ordered states, like the smectic.

On the other hand, mathematically we require that:

  1. $S$ to be continuous at every $\Omega$
  2. To be extensive with system size
  3. Differentiable (does it have to be always?)
  4. Path independent: state function
  5. ...what else?

Clearly if we don't know how to bound such functions based on a set of axioms, we cannot know if they will make sense physically and mathematically.

(Helpful thought scenario: Imagine studying a system of particles in suspension, where their positional distribution comes with a certain periodicity, if we are to attribute an entropic term to the state of periodicity of the system, what conditions should such state function satisfy.)

  • What are the axioms to be satisfied by a state function for it to qualify as entropy?
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Ultimate physical motivation

Strictly in the sense of physics, the entropy is less free than it might seem. It always has to provide a measure of energy released from a system not graspable by macroscopic parameters. I.e. it has to be subject to the relation $${\rm d}U = {\rm d}E_\text{macro} + T {\rm d} S$$ It has to carry all the forms of energy that cannot be expressible macroscopically, which we summarize as "heat" but the actual physics behind this "heat" might be quite different from the notions in gases etc. If entropy does not satisfy this relation, it is not a physical entropy. This would be a full characterization of entropy for macrophysics. I am going to use only this definition, not the cases where entropy is a handle to talk about information.


Statistical formulation

This constraint indeed does provide some freedom for the statistical definition of entropy, but not in effect. The freedom is basically in the fact that we are doing the $N\to \infty$ and $V \to \infty$ limits and a lot of information from the definition gets smeared out. We can for example define the phase space volume of the microcanonical ensemble in three distinct ways. First one is $$\Omega_\text{sharp} = \int_{\sum E = U} d \mu$$ Where $\mu$ is some kind of measure over the space of states. Or we can put $$\Omega_\text{non-sharp} = \int_{\sum E \in (U-\varepsilon,U)} d \mu$$ or even $$\Omega_\text{nobody cares} = \int_{\sum E < U} d \mu$$ Any of these will work for $S = k_B \log \Omega$ in the mentioned limits (the limit will give the same $S$). But this is more of a relict of the large limits - the physically plausible option is $\Omega_\text{sharp}$.

The much more important issue is counting the number of relevant states, the transition from discrete states to continuous ones and why we should consider them "democratic". This would be a very long argument involving ergodicity and so on.

For ergodic Hamiltonian systems, the probability measure is certainly proportional to $d^n x d^np$ where $n$ is the number of degrees of freedom. From quantum mechanics we know, that the "democracy" factor of discrete to continuous states makes this measure $d^n x d^np/h$ with $h$ the Planck constant. (Only the relative weights matter, since we normalize anyways.)

The conclusion is that the procedures of statistical physics, for a given system, can give us entropy unambiguously (up to an additive constant representing the freedom of state normalization).


Hand waivy conclusion

So there always is one entropy for every situation and we know how to derive it. The trick is only to specify which degrees are "free" or getting randomized in a complicated interaction and turn on the statistics.

But there are some loopholes. We see that the justification of the whole procedure (the "democratization" of states) relies on the Hamiltonian formulation and basically also quantization. But we know quantization is more of an art than a science and the statistical procedure can run into very similar problems as quantization. Are we always sure what the macroscopic parameters of a system are? How do we describe the situation when we observe the microstate directly? What would be the entropy of a relativistic space-time? Which would be the "activated" degrees of freedom? Etc. But this is a question for the "art of physics".


Additional note: "Art of physics" - modelling and confirming

A brief comment on "the art of physics". As with any physical models and approximations, there are three criteria:

  1. Foundation on (more) elementary physics
  2. Self-consistence of result with assumption
  3. Empirical verification

Say we have an open system $\Xi$ with a channel of particle inflow. However, we only know how to compute the parameters relevant for the inflow for small number densities in $\Xi$, because then we can use a one-particle model of entrance and leaving from the system. The one-particle model would be the point 1. - foundation on physics believed to be fundamental. We thus presume low number density and compute the statistics of the system.

But this is where the theorist's work should not stop, the last step is to check whether the density is sufficiently low under any choice of parameters and identify these regions in parameter space - this is point 2. However, this is a very primitive conception. For a serious model, the theorist should at least check whether two and higher particle models of inflow cannot suddenly take over even at low densities and investigate under what conditions they do not. This is 1. mixing with 2.

Nevertheless, there is also 3. - the empirical verification. It would be very naïve to pretend that the theorist is able to anticipate all the possible effects. In fact, Einstein's papers are well known to just shoot out a model without long mathematical discussions of neglected effects, and give experimental predictions right away. Sometimes, intuition rules (sometimes it also does not).

In the case of entropy this would be achieved by measuring the heat response of the system. It's not only heat capacities in the form $$C_{...} \sim \left(\frac{\partial S}{\partial T}\right)_{| ...\;=\text{const}.} $$ but also a lot of other response coefficient involving temperature as specified e.g. by the Maxwell relations.

So the answer would be: If a well developed model predicting quantitatively the entropy exists and it is confirmed by thorough testing, the entropy qualifies as the unique entropy of the system.


Additional note: Observed mathematical conditions

Let's say our the physical motivation is paramount. Then the strongest we can say is the following:

  • Entropy is a single-valued function of the full set of macroscopic parameters. (I.e. if it is not it might also be because the list of parameters is not complete.)
  • Entropy has a finite difference between any two points in the macro parameter space. I.e. $|\Delta S|<\infty$.
  • Entropy is homogeneous in the parameters defined by physical criteria as "extensive". I.e. for a complete set of extensive parameter $A_i$ we have $S(\lambda A_1, ...,\lambda A_n, ...) = \lambda S(A_1,...,A_n,...), \forall \lambda <0$.

In phase transitions as common as freezing/melting entropy is even discontinuous thus the criterion. (But this happens only in the $N \to \infty$ limit as discussed e.g. by Kardar in his notes.) Physically we are able to measure only $\Delta S$ so a strict requirement of well defined $dS$ is both redundant and impossible for some very common systems.

It is important that the "extensivity" is just saying "take another copy of the system" - the parameters which double by this operation are extensive but so is also heat stored in the new "double" system. Taking all the extensive parameters and multiplying by $\lambda$ just means "taking $\lambda$ copies of the system". This all relies heavily on the fact that we are able to very clearly identify the physical operation of "taking another copy of the system".

There are cases such as Black hole thermodynamics where such a notion fails. In a way, the whole space-time is the thermodynamical system, so "take another copy of the system" is hard to specify. (More technically, the formulas are for isolated black holes and there is no way to screen out gravity otherwise than by infinite distance.) It might seem that the horizon surface $A$ would be an extensive parameter but it actually grows as $\sim M^2$ - we cannot just say "double the mass" because that would not work.

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  • $\begingroup$ Thanks for your explanations. +1 for the structure and coherence of your answer. When you say "procedures of statistical physics...can give us entropy unambiguously" what do you mean exactly? For example one such procedure would be to obtain the partition function. But most of the time, in real systems, they're unsolvable (as I commented for BySymmetry's answer), so as you say, the trick is to boil down the problem to degrees of freedom and interaction that matter most and based on them we model these entropic terms, but what conditions account for their physical and mathematical validity? $\endgroup$ – Phonon Aug 29 '14 at 9:42
  • $\begingroup$ I tried to develop the point of "boiling down" briefly and added it to the answer, but I am not sure if this is what you are asking (Is it?). I guess you are also in a way trying to ask about how the statistical definition of entropy was pinned down by Boltzmann and others, but that would be an entirely different argument. $\endgroup$ – Void Aug 29 '14 at 14:58
  • $\begingroup$ Thanks for the further elaborations, very nice! I think you've summed it up pretty well in terms of physical requirements and steps in deriving relevant entropic terms in a general manner. Finally would you add anything to the mathematical conditions that are necessary in all such models of entropy? (only the ones that can be generalized of course, like the 4 points I had mentioned in the post), it would be "the cherry on top" of your answer as the saying goes :) $\endgroup$ – Phonon Aug 29 '14 at 21:02
  • $\begingroup$ The mathematical conditions are just derived from the original physical motivation. But once again physics is physics and any axioms can be bent, so I just stated the "strongest reasonable". I think this should answer your question sufficiently. $\endgroup$ – Void Aug 30 '14 at 9:23
  • $\begingroup$ Thanks a bunch for having added all the details, each time I asked for them. Honestly couldn't have asked for a more comprehensive answer, concise and coherent throughout. Cheers! One last question if I may, you say "in some phase transitions entropy is even discontinuous", but isn't it rather the derivatives of $S$ that should exhibit singularities for phase transition to occur? (e.g. singularity in the first order derivative $\rightarrow$ first order phase transition, etc) $\endgroup$ – Phonon Aug 30 '14 at 10:33
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[As requested, I convert my comment into an answer, as it might also be useful for other people.]

There is a very interesting series of works by Lieb and Yngvason on entropy and the second law of thermodynamics, based on the kind of axiomatic approach you seem to be interested in. You can start with this introductory paper, or this, this or this more detailed ones. There are also probably a few others by the same authors.

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  • $\begingroup$ It seems to me that somehow some of the axioms must kill phase transitions, as we cannot say "entropy is a concave and continuously differentiable function" due to them. (Theorem 5, Introductory paper) That is, these axiomatics seem to be built around a certain class of special systems. $\endgroup$ – Void Aug 30 '14 at 10:23
  • $\begingroup$ No. The approach is extremely general (much more so than previous ones). Concerning phase transitions, read the few lines just after the statement of the theorem you quote. $\endgroup$ – Yvan Velenik Aug 30 '14 at 11:00
  • $\begingroup$ Oh, I see, it is in the sense of "work coordinates" not the Legendre transforms, thanks. Then I guess it would be interesting to ask when handling an exotic thermodynamical system of which we can observe certain macroscopic observables and produced heat, whether we could say a set of observables is a complete set of work coordinates only based on the fact it takes no "latent" heats under such a description. $\endgroup$ – Void Aug 30 '14 at 13:35
  • $\begingroup$ Thanks again Yvan for the references, they are all very much in the line of what I was looking for, good stuff +1! $\endgroup$ – Phonon Aug 30 '14 at 14:10
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Thermodynamically entropy is defined by \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_{rev}}{T} \, ,\end{equation} where $\mathrm{d}Q_{rev}$ is the heat, transferred reversibly. As you point out it can be shown that this quantity is a function of state. This implies that the entropy of any thermodynamic system has, up to a constant, a well defined value given by \begin{equation} S = S_0 + \int_0^T \frac{C}{T^\prime}\mathrm{d}T^\prime \, ,\end{equation} where $C = \frac{\partial Q}{\partial T}$ is the heat capacity of the system. So if we set the value the entropy of a system at absolute zero, the laws of thermodynamics and the physical properties of the system fix the value of the entropy.

The formulae we have for the entropies of various systems are not, therefore, something that was derived rather than decided upon. Counting microstates is not a practical thing to do experimentally, but can be done theoretically for a given model. From this we can use Gibbs or Boltzmann's formulae to derive a formula for the entropy in terms of macroscopic quantities and this can be compared with experimental measurements of the heat capacity.

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    $\begingroup$ Thanks for your answer. In the last paragraph, you reach a point very close to what I'm asking, i.e. how entropies of various systems are "decided upon" rather than being "derived" from a common starting point. Which brings us to the question, ok although an element of intuition (based on empirical results maybe) is involved, one still has to give the entropy a mathematical model, what are the conditions any such entropic model should satisfy? (similar to the points I was using in the post). $\endgroup$ – Phonon Aug 28 '14 at 20:43
  • $\begingroup$ For any model the entropy is derived not decided upon. If you have a microscopic model of the system then you can straightforwardly write down the partition function. From this you can directly calculate the entropy. Statistical Mechanics gives a very general prescription for doing this. And as @CuriousOne rightly says this only allows you to calculate the the entropy of your model. The real world is always too complicated to talk about perfect mathematical axioms. $\endgroup$ – By Symmetry Aug 28 '14 at 21:27
  • $\begingroup$ Your points are fine and clear, it's just you still haven't touched upon the models themselves, afterall the main question is within what common framework do we define such entropic models? As for the partition function, I don't see what you mean by "straightforwardly write down...", because often one faces expressions with at least $6N$ integrals over positions and momenta of particles in the system, $H$ the Hamiltonian, then: $Z=\int d\mathbf{r}^N d\mathbf{p}^N \exp[H(\mathbf{r}^N,\mathbf{p}^N)/k_BT]$ Clearly we cannot always solve them analytically to deduce the expression of entropy. $\endgroup$ – Phonon Aug 29 '14 at 9:04
  • $\begingroup$ I was very careful about the placement of the word straightforward in that sentence. You can easily right down an expression for the partition function, you did it in your previous comment, but I never said it could be analytically evaluated in all cases. The point is that entropy has well defined value, whether we can write down an analytic expression for it or not. At other times we can use to numerical methods or find it empirically. In the cases where we do have an analytic expression it is normally because someone found a reasonable approximation for the relevant integrals. $\endgroup$ – By Symmetry Aug 29 '14 at 18:47

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